~ firearms and fermentation

Daily Archives: January 26, 2013

We were coated in a sheet of ice yesterday due to a winter storm so I decided to pull out an Eisbock brewed in 2002 by an old friend. The beer was a Silver medal in the 2002 AHA National Homebrew Competition and it had beer lurking in the back of my DBF ever since the brewer put it in my hot little hand. The style is not one that often pops up at location competition, but the methodology is to brew a doppelbock and concentrate the beer by freezing. This process brings out some really intense malt flavors and of course with water removed increases the level of alcohol. Over the years I have inadvertently frozen several beers and I can attest you do not want to do this with anything hoppy since those flavors become harsh. A nice smooth lager is the best beer to concentrate if you choose to do so and if it is legal to do so in your state.

What I thought might be interesting this morning would be to throw a little math in the mix and figure out the approximate alcohol content of the Eisbock. We’ll have to make some assumptions, but we’ll be in the ballpark. So the Eisbock I had last night had an original gravity (OG) of 1.085 and a finishing gravity (FG) of 1.018 before freezing and after removing the ice the FG was 1.022. For our first assumption we will have to assume the hydrometer was calibrated and correct.

Let’s initially assume 1 gallon of water was removed as ice from 5 gallons of beer. It’s pretty easy to figure the approximate alcohol of the original beer was (85 – 18) x 0.131 = 8.78%. Since alcohol is conserved in freezing then new ABV is pretty simple to calculate as well. (8.78 x 5) / 4 = 10.98%. So let’s check our math using that finishing gravity of the 5 gallon batch. 5 x 18 = 90 / 4 = 22.5 or 1.0225. So 1 gallon of ice could not have been removed since we did not end up with the same result. To end with the measured value of 1.022 we would need to end up with 5 x 18 / 22 = 4.09 gallons. So the actual amount water removed as ice was 5 – 4.09 = 0.91 gallon. Knowing this we can now calculate the actual ABV using the initial volume of 5 gallons and the final volume of 4.09 gallons. The equation is (8.78 x 5) / 4.09 = 10.73%. Ain’t math fun?

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